Atyendix VL 1097 perpendicular to the track. Find (i) the speed of the car, (2) the coefficient of friction if the car is just on the point of side-slipping. (Board of Education Exam. Stage 3, 1904.) Referring to Fig. 968. There are two forces, the weight of the car and the centrifugal force, and the resultant of these takes the same direction as the plumb-bob string, for their ratio is independent of weight, viz. :•- tan 22* • 2/2 _ tan 22° And V = ^.4.04 x 32 x 300 = 62*2 ft. per sec. or 42*4 miles per hr. Also the angle of resultant with the track = 12° .". fji = tan 12° = '208. Example 76. A Trarncar weighs 50 tons, and current is cut oft when the speed is 16 miles per hr. Reckoning" time from that instant, the following velocities and times were noted:— V (rniles per hr,) ... ......... 16 14 12 10 / (time in sees.)............ o 9*3 21 35 Calculate the average retarding force, and find the average-velocity from t = o to t = 35, Find distance travelled between these times. "If the law of resistance be F Ibs, = a 4- b V -f <; V2, find the values of a, t>, and<r, from the above observations. (Hons. Applied Mech. Exam.) Set out the velocities on a time base, Fig. 9680. area under V curve.' _ 129-5-+152-1 + J54 Average velocity - -......~Stffltoc""......... ~ ^ total time = i£44rniles Per = 50 x 224.0 Ibs. - f / - 32 •7 - 5ioo? where V is the decrease of velocity ia miles per hr. Hence average force,,* ' 874 Ibs.